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Physics 121 Quiz #1, January 8, 2016
Please show all work, thoughts and/or reasoning in order to receive partial credit. The quiz is
worth 10 points total.
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Suppose that you have two point charges where charge q1 = 1nC is located at
0.5,−0.1,−0.5 m and
another charge q2 = −9nC is located at −0.6,−0.7,−0.2 m .
1. What is the electric force on q1 due to q2 ?
! !
ro − rs
r̂ = ! ! =
ro − rs
=
!
F1,2
!
F1,2
0.5,−0.1,−0.5 m − −0.6,−0.7,−0.2 m
( 0.5 + 0.6 )2 + ( −0.1+ 0.7 )2 + ( −0.5 + 0.3)2 m
1.1,0.6,−0.3
= 0.854,0.466,−0.233
1.288
⎡ 1× 10 −9 C × ( −9 × 10 −9 C ) ⎤
q1q2
9 Nm 2
= k 2 r̂ = 9 × 10 C 2 ⎢
⎥ 0.854,0.466,−0.233
r1,2
(1.28m )2
⎢⎣
⎥⎦
= −4.17,−2.27,1.14 × 10 −8 N
2. What is the electric field at charge q1's location due to charge q2 ?
The electric field is given by:
!
! F1,2
−4.17,−2.27,1.14 × 10 −8 N
N
E=
=
= −41.7,−22.7,11.4
−9
q1
1× 10 C
C
3. If the charge q1 were released from rest at its initial location, the direction of q1's acceleration
would most likely be
a. directly toward the charge q2 .
b. directly away from the charge q2 .
c. not defined since the magnitude of q1's acceleration is zero and thus charge q1 would remain
at rest.
d. unable to be determined since the mass of q1 is unknown.
Physics121EquationSheet
ElectricForces,FieldsandPotentials
! !
!
QQ
r −r
F = k 1 2 2 r̂ ; r̂ = !o !s
r
ro − r
!
! F
E=
q
!
Q
EQ = k 2 r̂
r
QQ
PE = k 1 2
r
Q
V (r ) = k
r
ΔV
Ex = −
Δx
W = −qΔV f ,i
Constants
g = 9.8 sm2
1e = 1.6 ×10 −19 C
k=
1
= 9 ×109
4πε o
ε o = 8.85 ×10 −12
C2
Nm 2
Nm 2
C2
1eV = 1.6 ×10 −19 J
µ o = 4π ×10 −7 TmA
c = 3 ×108
m
s
h = 6.63 ×10 −34 Js
0.511MeV
c2
937.1MeV
m p = 1.67 ×10 − 27 kg =
c2
948.3MeV
− 27
mn = 1.69 ×10 kg =
c2
931.5MeV
1amu = 1.66 ×10 − 27 kg =
c2
N A = 6.02 ×10 23
me = 9.11×10 −31 kg =
Ax 2 + Bx + C = 0 → x =
− B ± B 2 − 4 AC
2A
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